Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space

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I've got ten (projective) planes in projective 3-space:



beginalign
&x=0\
&z=0\
&t=0\
&x+y=0\
&x-y=0\
&z+t=0\
&x-y-z=0\
&x+y+z=0\
&x-y+t=0\
&x+y-t=0
endalign

If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.



The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.










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    up vote
    7
    down vote

    favorite
    1












    I've got ten (projective) planes in projective 3-space:



    beginalign
    &x=0\
    &z=0\
    &t=0\
    &x+y=0\
    &x-y=0\
    &z+t=0\
    &x-y-z=0\
    &x+y+z=0\
    &x-y+t=0\
    &x+y-t=0
    endalign

    If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.



    The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.










    share|cite|improve this question























      up vote
      7
      down vote

      favorite
      1









      up vote
      7
      down vote

      favorite
      1






      1





      I've got ten (projective) planes in projective 3-space:



      beginalign
      &x=0\
      &z=0\
      &t=0\
      &x+y=0\
      &x-y=0\
      &z+t=0\
      &x-y-z=0\
      &x+y+z=0\
      &x-y+t=0\
      &x+y-t=0
      endalign

      If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.



      The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.










      share|cite|improve this question













      I've got ten (projective) planes in projective 3-space:



      beginalign
      &x=0\
      &z=0\
      &t=0\
      &x+y=0\
      &x-y=0\
      &z+t=0\
      &x-y-z=0\
      &x+y+z=0\
      &x-y+t=0\
      &x+y-t=0
      endalign

      If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.



      The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.







      co.combinatorics projective-geometry hyperplane-arrangements






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      asked 4 hours ago









      მამუკა ჯიბლაძე

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          This configuration has automorphisms by the symmetric group $S_5$,
          and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
          in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
          $$
          (x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
          $$

          the ten planes then have $(i,j)$ in the order
          $$
          (0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
          $$

          The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
          $a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
          Desargues
          configurations: intersecting with a generic plane $Pi$ in the 3-space
          yields the ten points and ten lines of a Desargues configuration,
          and any Desargues configuration can be obtained this way for some
          choice of $Pi$.






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            up vote
            4
            down vote













            This configuration has automorphisms by the symmetric group $S_5$,
            and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
            in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
            $$
            (x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
            $$

            the ten planes then have $(i,j)$ in the order
            $$
            (0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
            $$

            The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
            $a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
            Desargues
            configurations: intersecting with a generic plane $Pi$ in the 3-space
            yields the ten points and ten lines of a Desargues configuration,
            and any Desargues configuration can be obtained this way for some
            choice of $Pi$.






            share|cite|improve this answer
























              up vote
              4
              down vote













              This configuration has automorphisms by the symmetric group $S_5$,
              and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
              in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
              $$
              (x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
              $$

              the ten planes then have $(i,j)$ in the order
              $$
              (0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
              $$

              The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
              $a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
              Desargues
              configurations: intersecting with a generic plane $Pi$ in the 3-space
              yields the ten points and ten lines of a Desargues configuration,
              and any Desargues configuration can be obtained this way for some
              choice of $Pi$.






              share|cite|improve this answer






















                up vote
                4
                down vote










                up vote
                4
                down vote









                This configuration has automorphisms by the symmetric group $S_5$,
                and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
                in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
                $$
                (x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
                $$

                the ten planes then have $(i,j)$ in the order
                $$
                (0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
                $$

                The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
                $a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
                Desargues
                configurations: intersecting with a generic plane $Pi$ in the 3-space
                yields the ten points and ten lines of a Desargues configuration,
                and any Desargues configuration can be obtained this way for some
                choice of $Pi$.






                share|cite|improve this answer












                This configuration has automorphisms by the symmetric group $S_5$,
                and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
                in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
                $$
                (x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
                $$

                the ten planes then have $(i,j)$ in the order
                $$
                (0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
                $$

                The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
                $a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
                Desargues
                configurations: intersecting with a generic plane $Pi$ in the 3-space
                yields the ten points and ten lines of a Desargues configuration,
                and any Desargues configuration can be obtained this way for some
                choice of $Pi$.







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                answered 33 mins ago









                Noam D. Elkies

                55.2k10195280




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